Linear Differential Operators - Penn Engineeringdorny/VectorSp/diffoperatorsp93-143.pdf · The Role...

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Linear Differential Operators

Differential equations seem to be well suited as models for systems. Thusan understanding of differential equations is at least as important as anunderstanding of matrix equations. In Section 1.5 we inverted matrices andsolved matrix equations. In this chapter we explore the analogous inversionand solution process for linear differential equations.

Because of the presence of boundary conditions, the process of invertinga differential operator is somewhat more complex than the analogousmatrix inversion. The notation ordinarily used for the study of differentialequations is designed for easy handling of boundary conditions rather thanfor understanding of differential operators. As a consequence, the conceptof the inverse of a differential operator is not widely understood amongengineers. The approach we use in this chapter is one that draws a stronganalogy between linear differential equations and matrix equations,thereby placing both these types of models in the same conceptual frame-work. The key concept is the Green’s function. It plays the same role for alinear differential equation as does the inverse matrix for a matrix equa-tion.

There are both practical and theoretical reasons for examining theprocess of inverting differential operators. The inverse (or integral form) ofa differential equation displays explicitly the input-output relationship ofthe system. Furthermore, integral operators are computationally andtheoretically less troublesome than differential operators; for example,differentiation emphasizes data errors, whereas integration averages them.Consequently, the theoretical justification for applying many of the com-putational procedures of later chapters to differential systems is based onthe inverse (or integral) description of the system. Finally, the applicationof the optimization techniques of Chapters 6-8 to differential systems oftendepends upon the prior determination of the integral forms of the systems.

One of the reasons that matrix equations are widely used is that we havea practical, automatable scheme, Gaussian elimination, for inverting amatrix or solving a matrix equation. It is also possible to invert certaintypes of differential equations by computer automation. The greatestprogress in understanding and automation has been made for linear,

93

94 Linear Differential Operators

constant-coefficient differential equations with initial conditions. Theseequations are good models for many dynamic systems (systems whichevolve with time). In Section 3.4 we examine these linear constant-coefficient models in state-space form and also in the form of nth-orderdifferential equations. The inversion concept can be extended to partialdifferential equations.

3.1 A Differential Operator and Its Inverse

Within the process of inverting a differential operator there is an analogueof the elimination technique for matrix inversion. However, the analogybetween the matrix equation and the differential equation is clouded bythe presence of the boundary conditions. As an example of a lineardifferential equation and its associated boundary conditions, we use

-f” =u w i t h f(O)= art a n d f(b) = CQ (3.1)

Equation (3.1) can be viewed as a description of the relationship betweenthe steady-state temperature distribution and the sources of heat in aninsulated bar of length b. The temperature distribution f varies only as afunction of position t along the bar. The temperature distribution iscontrolled partly by u, the heat generated (say, by induction heating)throughout the bar, and partly by constant temperature baths (oftemperatures cyr and [x2, respectively) at the two ends t = 0 and t = b. Thusboth the distributed input u and the boundary inputs { ai} have practicalsignificance. The concepts of distributed and boundary inputs extend toother ordinary and partial differential equations.

A Discrete Approximation of the Differential System

In order to obtain a more transparent analogy to matrix equations andthereby clarify the role of the boundary conditions, we temporarilyapproximate the differential equation by a set of difference equations.* Letb = 4, substitute into (3.1) the finite-difference approximation

*The approximation of derivatives by finite differences is a practical numerical approach tothe solution of ordinary and partial differential equations. The error owing to the finite-difference approximation can be made as small as desired by using a sufficiently fineapproximation to the derivatives. (See Forsythe and Wasow [3.3].) Special techniques areusually used to solve the resulting algebraic equations. See P&C 3.3 and Varga [3.12].

Sec. 3.1 A Differential Operator and Its Inverse

and evaluate the equation at t = 1, 2, and 3:

95

-f(O)+2f(l) -f(2) =u(l)-f(l)+2f(2) -f(3) = u(2)

- f(2) + 2f(3) - f(4) = u(3) (3.4)

f(O) =a

f(4) =a:

It is obvious that this set of algebraic equations would not be invertiblewithout the boundary conditions. We can view the boundary conditionseither as an increase in the number of equations or as a decrease in thenumber of unknowns. The left side of (3.2), including the boundaryconditions, is a matrix multiplication of the general vector (f(0)f( 1) * ’ * f(4))= in the space %5x ‘. The corresponding right-hand side of(3.2) is (u(1) u(2) u(3) (xi (~2)~; the boundary values increase the dimensionof the range of definition by two. On the other hand, if we use theboundary conditions to eliminate two variables, we reduce the dimensionof the domain of the matrix operator by two, f(0) and f(4) become part ofthe right-hand side, and the reduced matrix operates on the general vector(f(l) f(2) f(3))’ in 9R,3x ‘. By either the “expanded” or the “reduced” view,the transformation with its boundary conditions is invertible. In the nextsection we explore the differential equation and its boundary conditionsalong the same lines as we have used for this discrete approximation.

The Role of the Boundary Conditions

A differential operator without boundary conditions is like a matrix withfewer rows than columns: it leads to an underdetermined differentialequation. In the same manner as in the discrete approximation (3.2),appropriate boundary conditions make a linear differential operator invert-ible. In order that we be able to denote the inverse of (3.1) in a simplemanner as we do for matrix equations, we must combine the differentialoperator - D2 and the two boundary conditions into a single operator on avector space. We can do so using the “increased equations” view of theboundary conditions. Let f be a function in the space e2(0,b) of twicecontinuously differentiable functions; then -f” will be in C?(O,b), thespace of continuous functions. Define the differential system operator T:e2(0, b)-+ e (0, b) x ‘JR2 by

Tf 9 (-f”,f(O),f(b)) (3.3)

The system equations become

Tf = (u, “1,(x2) (3.4)


We are seeking an explicit expression of T- ’ such that f = T- ‘(u, ai, (w,).Because of the abstractness of T, an operation which produces a mixtureof a distributed quantity u and discrete quantities {(xi}, it is not clear howto proceed to determine T-l.

Standard techniques for solution of differential equations are moreconsistent with the “decreased unknowns” interpretation of the boundaryconditions. Ordinarily, we solve the differential equation, - f” = u, ignoringthe boundary conditions. Then we apply the boundary conditions toeliminate the arbitrary constants in the solution. If we think of theoperator - D2 as being restricted through the whole solution process to actonly on functions which satisfy the boundary conditions, then the“arbitrary” constants in the solution to the differential equation are notarbitrary; rather, they are specific (but unknown) functions of theboundary values, { aj}. We develop this interpretation of the inversionprocess into an explicit expression for the inverse of the operator T of (3.3).

How do we express the “restriction” of -D2 in terms of an operator ona vector space? The set of functions which satisfy the boundary conditionsis not a subspace of e2(0,b); it does not include the zero function (unlessa =a1 2=O). The analogue of this set of functions in the three-dimensionalarrow space is a plane which does not pass through the origin. We framethe problem in terms of vector space concepts by separating the effects ofthe distributed and boundary inputs. In point of fact, it is the difference inthe nature of these two types of inputs that has prevented the differentialequation and the boundary conditions from being expressed as a singleequation.* Decompose the differential system (3.1) into two parts, oneinvolving only the distributed input, the other only the boundary inputs:

-fpu with f,(O) = fd( b) = 0 (3.5)

-fp(j with f,(O) = LYE, fb( b) = a2 (3.6)

Equations (3.5) and (3.6) possess unique solutions. By superposition, thesesolutions combine to yield the unique solution f to (3.1); that is, f = fd+ fb.Each of these differential systems can be expressed as a single operator ona vector space. We invert the two systems separately.

We work first with (3.5). The operator -D2 is onto C?(O, b); that is, wecan obtain any continuous function by twice differentiating some functionin e2(0, b). However, -D2 is singular; the general vector in nullspace(- D2) is of the form f(t) = ci + c,t. We modify the definition of theoperator - D2 by reducing its domain. Let ?f be the subspace of functionsin e2(0,b) which satisfy the homogeneous boundary conditions of (3.5),

*Friedman [3.4] does include the boundary conditions in the differential equation by treatingthe boundary conditions as delta functions superimposed on the distributed input.

Sec. 3.1 A Differential Operator and Its Inverse 97

f(0) = f(b) = 0. Define the modified differential operator Td: y+ (? (0, b) by

Tdf k -D2f for all f in V. The “distributed input” differential system (3.5)becomes

(3.7)

The boundary conditions are now included in the definition of theoperator; in effect, we have “reduced” the operator -D2 to the operatorT, by using the two boundary conditions to eliminate two “variables” ortwo degrees of freedom from the domain of the operator -D2. Theoperator Td is nonsingular; the equation - f”(t) = 0 has no nonzerosolutions in Ir. Furthermore, T, is onto; eliminating from the domain of- D2 those functions which do not satisfy the zero boundary conditions of(3.5) does not eliminate any functions from the range of -D2. Suppose g isin e2(0, b), and that g(0) and g(b) are not zero. Define the related function

f in ‘v by f(t) i g(t)-[g(O)+ t(g(b)-g(O))/b]. We have simply subtracteda “straight line” to remove the nonzero end points from g; as a result,f(O)=f(b)=O. But -D2f = - D’g. Both f and g lead to the same function inC?(O,b). Every vector in C?,(O, b) comes (via - D2) from some function inY. Thus Td is onto and invertible.

The differential system (3.6) can also be expressed as a single invertibleoperator. The nonzero boundary condit ions of (3.6) describe atransformation U: C2(0, b)+ CR,‘, where

Uf i (f(O),f(b))

Since e2(0, b) is infinite dimensional but CR2 is not, U must be singular. Wemodify the operator U by reducing its domain. Let % be the subspace offunctions in e2(0,b) which satisfy the homogeneous differential equationof (3.6), -f;(t) = 0; %J is the two-dimensional space Y2 consisting infunctions of the form f(t)= ci + c,t. We define the modified operator T,:

?I)‘+ a2 by T,f e (f(O), f(b)) for all f in 9’. The “boundary input” differen-tial system (3.6) can be expressed as the two-dimensional equation

(3.8)

The differential equation and boundary conditions of (3.6) have beencombined into the single operator, T,. I t is apparent that T, isinvertible-the operator equation is easily solved for its unique solution.

The Inverse Operator

We have rephrased (3.5) and (3.6) in terms of the invertible operators T,and T,, respectively. Because (3.5) and (3.6) constitute a restructuring of


(3.1), we can express the solution to (3.1) as

=T-‘(u,ar,,a2) (3.9)

where T is the operator of (3.3).Since T, is a differential operator, we expect T,’ : e(O, b)-+ ‘v to be an

integral operator. We express it explicitly in the general form (2.34):

fd(t)=(T,‘u)(t)=/bk(t,s)u(s)ds (3.10)0

The kernel function k is commonly referred to as the Green’s function forthe differential system (3.1). In order that (3.10) correctly express theinverse of T,, fd(t) must satisfy the differential system (3.5) from which T,is derived. Substituting (3.10) into (3.5) yields

d2 b-f;(t)= - --& k(t,s)u(s)ds

Jb d2k(w)

= -

dt2u(s)h=u(t)

0

with

fd(0) = lbk(O,s)u(s)ds = 00

f,(b)= &=‘k(b,s)u(s)ds=O

for all u in (Z (0, b). These equations are satisfied for all continuous u if andonly if

d2k(t,s)-

dt2=qt-s) with k(O,s) = k(b,s) =0 (3.11)

That is, the Green’s function k, as a function of its first variable t, mustsatisfy the differential equation and boundary conditions (3.5) for u(t)=8(t - s), where 6 (t-s) is a unit impulse (or Dirac delta function) appliedat the point t = s.* We can use (3.11) to determine the Green’s function.

*See Appendix 2 for a discussion of delta functions. We use some license in interchanging theorder of differentiation and integration when delta functions are present. The interchange canbe justified, however, through the theory of distributions (Schwartz [3.10]).


For practical purposes we can think of 6 (t - s) as a narrow continuouspulse of unit area, centered at t = s. [In terms of the steady-state heat-flowproblem (3.1), the function 6 (t - s) in (3.11) represents the generation of aunit quantity of heat per unit time in the cross section of the bar at t = s.]However, 6 (t - s) is not a function in the usual sense; its value is notdefined at t = s. It is not in e2(0, b). Therefore, the solution k to (3.11)cannot be in C2(0, b). We simply note that the domain e2(0, b) and rangeof definition C?(O, b) of the operator - D2 were defined somewhatarbitrarily. We can allow a “few” discontinuities or delta functions in- D2f if we also add to e2(0, b) those functions whose second derivativescontain a “few” discontinuities or delta functions.

The operator T, - ’ : 9L2+ 9’ can also be expressed explicitly. Since T, - ’acts linearly on the vector (LX,, 1x2) in CR2 to yield a polynomial in Y2, weexpress T, - ’ as

fb=Tb-l(al,a2)=a1P,+a2P2 (3.12)

where p1 and p2 are functions in ‘?? 2. We refer to the function pj(t) as theboundary kernel for the differential system (3.1). Just as the Green’sfunction is a function of two variables, t and s, so the boundary kernel is afunction of both the continuous variable t and the discrete variable j.Because of the simplicity of the differential operator of this example, theintroduction of the boundary kernel seems unnecessary and artificial. Formore complicated differential operators, however, the boundary kernelprovides a straightforward approach to determination of the full inverseoperator. In order that (3.12) correctly describe T,-‘, f, must satisfy thedifferential system (3.6):

-fb” = - qp;’ - a2p; = 0

f,(O) = ‘yIPI + ‘x2p2P) = a1

f,(b) = 'YIP@) + a2~2@> = a2

for all (x1 and (Ye. Thus the boundary kernel p must obey

-p;‘(t)=0 w i t h p,(O)=l, p,(b)=0

-pi(t)=0 w i t h p2(0)=0, p2(b)= 1(3.13)

We can use (3.13) to determine the boundary kernel.We have defined carefully the differential system operator T, the “dis-

tributed input” system operator T,, and the “boundary input” systemoperator T, in order to be precise about the vector space concepts involvedwith inversion of differential equations. However, to continue use of this


precise notation would require an awkward transition back and forthbetween the vector space notation and the notation standard to the field ofdifferential equations. We rely primarily on the standard notation. We usethe term differential system to refer to the differential operator with itsboundary conditions (denoted {- D2, f(O), f(b)} in this example) and alsoto the differential equation with its boundary conditions [denoted as in(3.1)]. We refer to both the inverse of the operator and the inverse of theequation as the inverse of the differential system. Where we refer to thepurely differential part of the system separately, we usually denote itexplicitly, for example, as - D2 or as - f” = u.

A Green’s Function and Boundary Kernel

We solve for the Green’s function k of the system (3.1) by direct integra-tion of (3.11). The successive integration steps are depicted graphically inFigure 3.1. It is clear from the figure that the integral of - d2k/ dt2 isconstant for t <s and t >s, and contains a jump of size 1 at t=s. Wepermit the value of the constant c to depend upon the point s at which theunit impulse is applied.

dk(t,s)- - = c(s),

dtt<s

=c(s)+ 1, t>s

Integration of - dk/dt yields continuity of - k at s:

-k(t,s)=c(s)t+d(s), t<s

=c(s)s+d(s)+(c(s)+l)(t-s), t>s

Applying the boundary conditions we find

- k(O,s)=c(s)(O)+d(s)=O *d(s)=0

- k(b,s)=c(s)s+(c(s)+ l)(b-s)=O a(s)= 9

Thus

(b-s)tk(td b=-, t<s

(b - t)s=-, t>s

b(3.14)


Figure 3.1. Graphical integration of (3.11).

where both t and s lie in the interval [0, b].By integration of (3.13) we determine the boundary kernel p associated

with (3.1). The general solution to the jth differential equation is p,.(t)=Cjl+ c,+. Using the boundary conditions we find

(3.15)

Having found k and p, we insert them into (3.10) and (3.12) to obtain


T,-’ and T,-‘. Combining the two inverses as in (3.9) produces

(3.16)

Equation (3.16) is an explicit description of the inverse of the lineardifferential system (3.1).

A Matrix Analogy

A differential equation with an appropriate set of boundary conditions isanalogous to a square matrix equation. We explore this analogy in order toremove some of the abstractness and mystery from differential operatorsand their inverses. An example of a matrix equation and its correspondinginverse is

c :)(:;)=(~3 and (::)=L: -Xi:)Any such pair of equations can be expressed as Ax= y and x= A-‘y,respectively, for some square matrix A. The inverse matrix equation ismore clearly analogous to an inverse differential equation (or integralequation) if we express the matrix multiplication in the form of asummation. Denote the elements of x and y by & and Q, respectively. Thenthe equation x = A- ‘y becomes*

&= ,$ (A-‘)~B+ i = I,..., n (3.17)

The symbol (A-‘), represents the element in row i and column j of then X n matrix A- ‘. Thus the inverse matrix, a function of the two integervariables i and j, is the kernel of a summation operator. In the form (3.17),the inverse matrix equation x =A- ‘y is obviously a discrete analogue ofthe integral equation fJt)=J$(d,s)u(S)ds of (3.10). The Green’s functionk( t,s) is the analogue of the inverse matrix A-‘. If we compare the inversematrix equation (3.17) to (3.16), the full inverse of the differential system(3.1), the analogy is clouded somewhat by the presence of the boundaryterms. The true analogue of A-’ is the pair of kernel functions, k and p.

*See (2.35).

Sec. 3.2 Properties of nth-Order System and Green’s Functions 103

Because k(t,s) and Pi(t) appear as “weights” in an integral or summation,the inverse form of the differential system is somewhat more useful to theintuition than is the differential system itself.

We can also draw an analogy between the process of inverting thematrix A and the process of solving for k and p. The solution to theequation Ax = Ei, where ei is the ith standard basis vector for 9Lnx ‘, is theith column of A- l. The solution process is analogous to solving (3.11) fork(t,s) with s fixed; it is also analogous to solving (3.13) for b with j fixed.The row reduction (A i I)+(1 i A-‘) produces all columns of the inversematrix simultaneously. Thus the inversion of A by row reduction isanalogous to the determination of k and p by solving (3.11) and (3.13),respectively. In general, the process of computing k and p requires moreeffort than does the direct solution of (3.1) for specific inputs u and { ai}.However, the resulting inverse equation (3.16) contains information aboutthe solution for any set of inputs.

3.2 Properties of nth-Order Systems and Green’s Functions

In Section 3.1 we introduced the concepts of a differential operator and itsinverse by means of a simple second-order example, (3.1). We now explorethese concepts in detail for more general linear differential systems.Included in this section is an examination of noninvertible differentialsystems and a development of conditions for invertibility. Techniques forexplicit determination of the Green’s function and boundary kernel aretreated in Section 3.3.

We define a regular nth-order linear differential operator L : P (a, b)+e(G) by

(Lf)(t) Ai go(t)~“‘(t)+gl(t)f’“-“(t)+ l ’ * +g,(t)f(t) (3.18)

where the coefficients { g,.} are continuous and go( t)#O on [a, b].* Thecorresponding nth-order differential equation is Lf = u, where the distri-buted input function u is continuous on [a, b]. It is well known that L iso n t o C?(a,b); t h e nth-order differential equation without boundary condi-tions always has solutions (Ince [3.6]). The homogeneous differential equa-tion is defined as the equation Lf = 0, without boundary conditions (theinput u is zero). The homogeneous differential equation for the operator

*If the interval [a,b] were infinite, if go were zero at some point, or if one of the coefficientfunctions were discontinuous, we would refer to (3.18) as a singular differential operator. InSection 5.5 we refer to the regular second-order linear differential operator as a regularSturm-Liouville operator.


(3.18) always has n linearly independent solutions†; we call a set {v,, . . . , vn}of independent solutions a fundamental set of solutions for L. We some-times express such a set as the complementary function for L:

(3.19)

where c,,..., cn are unspecified constants. Both the complementary func-tion f, and the fundamental set of solutions {vi} are, in reality, descriptionsof the n-dimensional nullspace of L.

In order that L of (3.18) be invertible, we must add n appropriateboundary conditions to eliminate the n arbitrary constants in the com-plementary function. We denote the ith boundary condition for (3.18) by&(f) = a,., where ai is a scalar and csi is a linear functional on (?” (a, b).g Atypical boundary condition is some linear combination of f and its firstn - 1 derivatives evaluated at the end points of the interval of definition.For example,

PI(f) f y,f(a) + y,f’(a) + 7$(b) + YJV) = (~1 (3.20)

(where the { yi} are scalars) is as general a boundary condition as we wouldnormally expect to encounter for a second-order differential operatoracting on functions defined over [a,b]. The second boundary condition forthe second-order differential equation, &(f) = (Ye, would be of the sameform, although the particular linear combination of derivatives whichconstitutes p2 would have to be linearly independent of that specified bythe coefficients (y ,, y2, ys, y4) in &. There is, of course, no reason why theboundary conditions could not involve evaluations of f and its derivativesat interior points of the interval of definition. We refer to the boundarycondition 1si(f) = 0, where the boundary input q is zero, as a homogeneousboundary condition.

Consider the following nth-order differential system:

Lf=u

iSi(f)=cwi, l,...,mi =(3.21)

where L is defined in (3.18) and Isi is an n th-order version of (3.20); m istypically but not necessarily equal to n. We call a solution fP to (3.21) a

†See P&C 3.4.‡Of course, it is possible for the boundary conditions associated with a physical system to benonlinear functions of f. We consider here only linear differential equations and linearboundary conditions.

Sec. 3.2 Properties of nth-Order Systems and Green’s Functions 105

particular solution for the differential system. A completely homogeneoussolution fh for the differential system is a solution to the homogeneousdifferential equation with homogeneous boundary conditions (the homo-geneous differential system):

Lf=8 (3.22)&(f)=O, l,...,mi =

Thus a completely homogeneous solution for the differential system is asolution with all inputs zero. Any solution f to (3.21) can be written asf = fp + fh, where fP is any particular solution and f, is some homogeneoussolution. The set of completely homogeneous solutions constitutes thenullspace of the differential system (or the nullspace of the underlyingdifferential operator).* A system with a nonzero nullspace is not invertible.

Exercise 1. Suppose

(Lf)(t) i f”(t)=u(t) (3.23)

with the boundary conditions

P,(f) i fyO)= a1 /s2(f) i f’(1) = c9 (3.24)

What is the completely homogeneous solution to (3.23)-(3.24)? Show thatthe general solution to (3.23)-(3.24) is

f(t)=S,‘i&)d7do+a,t+f(O) (3.25)

where

I1u(T)dT= al - a2

0(3.26)

Note that the differential system (3.23)-(3.24) is not invertible. No solutionexists unless the inputs u and { cyi} satisfy (3.26).

The Role of the Homogeneous Differential System

The matrix analogue of the nth-order differential system (3.21) is thematrix equation Ax= y (where A is not necessarily square). Row reductionof A determines the nullspace of A (the solution to Ax= 0); it also shows

*See (3.4) and (3.9).


the dependencies in the rows of A and the degree of degeneracy of theequation-the degree to which the range of the matrix transformation failsto fill the range of definition. To actually find the range of the matrixtransformation (specific conditions on y for which the equation is solv-able), we can follow either of two approaches: (a) row reduce AT (the rowsof AT span the range of A)*; or (b) row reduce (A i I). If A is square andinvertible, approach (b) amounts to inversion of A.

For the differential system (3.21), the analogue of row reduction of A isthe analysis of the completely homogeneous system (3.22). We focus firston this analysis, thereby determining the extent to which (3.21) is under-determined or overdetermined. Then assuming the system (3.21) is invert-ible, we perform the analogue of row reduction of —inversion of thedifferential operator.

The solutions to the homogeneous differential equation, Lf = 8, areexpressed as the complementary function f, of (3.19). We apply the mhomogeneous boundary conditions to f,, thereby eliminating some of thearbitrary constants in f, :

(3.27)

The nullspace of the differential system (3.21) consists in the functionsf, = CIVl + * l l + c,v,, where some of the arbitrary constants {ci} areeliminated by (3.27).

The key to the differential system (3.21) lies in B, a boundary conditionmatrix (or compatibility matrix) for the system. In point of fact, B com-pletely characterizes (3.22). It describes not just the boundary conditions,but rather the effect of the boundary conditions on a set of fundamentalsolutions for L. In general, the m boundary conditions, in concert with the

*see P&C 2.19. In Section 5.4 we introduce the adjoint operator, the analogue of AT. Theorthogonal decomposition theorem (5.67) is the basis of a method for determining the range

AT.of an operator from the nullspace of its adjoint; this method is the analogue row reductionof


nth-order differential equation Lf = u, can specify either an underdeter-mined or overdetermined set of equations. Exercise 1 exhibits symptoms ofboth the underdetermined and overdetermined cases. Of course, B is notunique since it can be based on any fundamental set of solutions. Yet therank of B is unique; rank(B) tells much about the solutions to (3.21)*:

1. If rank(B) = m = n, then (3.27) precisely eliminates the completelyhomogeneous solution, and (3.21) is then the analogue of an invertiblesquare matrix equation; the system is invertible.

2 . I f rank(B) = p < n , then (n - p ) of the constants { ci} in thecharacteristic function remain arbitrary and the nullspace of the systemhas dimension (n - p). There are (n - p) degrees of freedom in the solutionsto (3.21); the system is singular.

3. If rank(B) = p < m, then (m - p ) rows of B are dependent on the rest.As demonstrated by Exercise 1, these dependencies in the rows of B mustbe matched by (m - p) scalar-valued relations among the boundary values{ ai} and the distributed input u, or there can be no solutions to (3.21)(P&C 3.5). The system is not onto e(a, b).

The following example demonstrates the relationship between rank(B) andthe properties of the differential system.

Example 1. The Rank of the Boundary Condition Matrix. Let

(M)(t) f f”(t)=u(t)

for t in [0, 1]. The set {vt,v2}, where VI(t)= 1 and vz(t)= t, is a fundamental set ofsolutions for t. We apply several different sets of boundary conditions, demonstrat-ing the three cases mentioned above.

1. #4(r) 4 w-3 = al, 820 b f(l) = CQ. In this case,

Since rank(B) = 2 = m = n, the system is invertible. We find the unique solution bydirect integration:

2 . &(f) i f(O)= (pi. For this single boundary condition,

B=(v,(O) vdO))=(l 0)

*Ince [3.6].


and rank(B) = 1. Since n = 2, we should expect one degree of freedom in thesolution. Since m = rank(B), we should expect a solution to exist for all scalars a,and all continuous functions u. By direct integration, the solution is

f(t)=Ifl~~(~)~~rlr+d,t+a,0 0

where di is an arbitrary constant.

3. &(f) 9 f(0) = al, &(f) 4 f( 1) = a2, &(f) f f’(0) = a3. Then,

Because rank(B) = 2 and m = 3, one scalar-valued function of u, al, a2, and a3 mustbe satisfied in order that a solution exist. Since rank(B)= n, if a solution exists for agiven set of inputs (u, al, a2), that solution is unique. We find the solution by directintegration and application of the three boundary conditions:

where u, al, a2, and a3 must satisfy

a2-al-a3-//

’ su(7)dds=o0 0

4 . B,(f) i f’(0)= al, 132(f) b f’(l) = a2. This case is presented in Exercise 1.

Rank(B) = 1, but m = n = 2. We expect one scalar-valued condition on the inputs,and one degree of freedom in the solutions. The general solution and the restrictionon the inputs are given in (3.25) and (3.26), respectively.

It is apparent from Example 1 that if m < n, the system is underdeter-mined; there are at least (n - m ) degrees of freedom in the solutions. Onthe other hand, if m > n, the system is usually overdetermined; sincerank(B) < n for an nth-order differential system, the input data must satisfyat least (m - n ) different scalar-valued restrictions in order that thedifferential equation and boundary conditions be solvable.

Ordinarily, m = n; that is, the differential equation which represents aphysical system usually has associated with it n independent boundary


conditions { #Ii}. These n boundary conditions are independent in the sensethat they represent independent linear combinations of f, f(l), . . . , f(“- ‘)evaluated at one or more points of [a,b]. However, we see from the fourthcase of Example 1 that a boundary condition matrix B can be degenerateeven if the boundary conditions are independent. Thus for a “square”differential operator, the condition for invertibility (or compatibility) is

det(B) # 0 (3.28)

where B is a boundary condition matrix as defined in (3.27).It can be shown that (3.28) is satisfied for any differential operator for

which m = n and for which the boundary conditions are linearly indepen-dent and are all at one point (P&C 3.4). Only for multipoint boundaryvalue problems can the test (3.28) fail. Exercise 1 is such a case.

For the rest of this chapter we assume (3.28) is satisfied, and proceed todetermine the inverse of the differential system (3.21). In Section 4.3,where we determine eigenvalues and eigenfunctions of differential opera-tors, we seek conditions under which (3.28) is not satisfied. These condi-tions occur, of course, only with multipoint boundary value problems.

The Green’s Function and the Boundary Kernel

Our procedure for inverting the system (3.21) parallels the procedure usedwith the second-order example (3.1). Of course, the compatibility condition(3.28) must be satisfied. Assume m = n. We begin by splitting (3.21) intotwo parts, one involving only the distributed input, the other only theboundary inputs:

Lf=uIsi( i=l,...,n (3.29)

Lf=e&(f)=q i= l,...,n (3.30)

where L is given in (3.18). The completely homogeneous equation (3.22) isa special case of both (3.29) and (3.30). Thus both are characterized by anyboundary condition matrix B derived from (3.22). If (3.28) is satisfied, both(3.29) and (3.30) are invertible. The inverse of (3.29) is an integral operatorwith a distributed kernel. The inverse of (3.30) is a summation operatorinvolving a boundary kernel. These two kernels describe explicitly thedependence of f(t) on the input data u(t) and { oli).

Assume the inverse of (3.29) is representable in the integral form

(3.31)


for all t in {a,b}. The kernel k is known as the Green’s function for thesystem (3.21). If (3.31) is the correct inverse for (3.29), f must satisfy (3.29):

(Lf)(t)=Llbk(t,s)u(s)a5a

= ‘Lk(t,s)u(s)ds=u(t)sa

( Wf)(t)-Wjbk(t,s)u(s)~a

= b&.k(t,s)u(s)ds=O/

j = l,...,na

for all u in C?(a,b). Both L and pi treat the variable s as a constant, actingon k(t,s) only as a function of t. Each operator acts on the whole “t”function k(-,s). It is evident that Lk(t,s) exhibits the “sifting” property ofa delta function (see Appendix 2). On the other hand, ~ik(t,S) acts like thezero function. Consequently, the Green’s function k must satisfy

d*k( t,s)Lk(t,s) ’ go(t) dtn + * *. +gn(t)k(t,s)=a(t-s)

(3.32)~ik(t,s)=O l,...,nj =

for all t and s in [a,b]. Because the delta function appears in (3.32), wecannot rigorously interchange the order of the differential operator L andthe integration without resorting to the theory of generalized functions(Appendix 2). However, we can justify the formal interchange for eachspecific problem by showing that the Green’s function k derived from(3.32) does indeed lead to the solution of (3.29) for every continuousfunction u.

Assume the inverse of (3.30) is representable as a summation operator ofthe form:

f(t)= i Pj(t)ajj-1

(3.33)

We can think of p as a kernel function of the two variables j and t. We callp the boundary kernel for (3.21). To find the equations which determine p,


we substitute (3.33) into (3.30):

Lf=L ~ Pj~jj=l

n= C aj(LPj)=e

j-1

= i ajP,(pj)=ai i=l,...,nj- 1

for all { ai}. Suppose we let ak = 1 and aj = 0 for j# k. It follows that fork=l ,...,n, Lp,=O, 1sipk’l for k=i, and flip,=0 for k#i. Thus theboundary kernel p must satisfy

dnPj( t,(LPj)(t) ’ go(t)7 + * * l +g*(t)&(t)=o j= L..,n (3.34)

isi (pi) = ag mj = 1 9*-*, n; j= l,...,n

for all t in [a, b], where 60 is the Kronecker delta (see A2.11 of Appendix2). According to (3.34), the n components {pi) of the boundary kernelconstitute a fundamental set of solutions for the operator L; furthermore,{pi) is a fundamental set for which the boundary condition matrix B of(3.27) is the n X n identity matrix.

By solving (3.32) and (3.34), we can invert any regular n th-orderdifferential system which has a nonsingular boundary condition matrix.The inverse of the differential system (3.21) (with m = n) consists in thesum of the inverses of (3.29) and (3.30), namely,

f(t)= Jbk(t,s)u(s)l+ 5 ajpj(t)a j-1

(3.35)

where k and p are determined by (3.32) and (3.34), respectively.Theoretically, we can invert any linear differential operator, ordinary or

partial, which has appropriate boundary conditions. That is, we canconvert any invertible linear differential equation to an integral equationanalogous to (3.35). As a model for a system, the integral equation is more

I12 Linear Differential Operators

desirable than the differential equation from two standpoints. First, theboundary conditions are included automatically. Second, integral opera-tors tend to “smooth” functions whereas differential operators introducediscontinuities and delta functions.* It is well known that numericaldifferentiation amplifies errors in empirical data, but numerical integrationdoes not (Ralston [3.9, p. 791). The rest of this chapter is devoted totechniques for determining the inverse (or integral) model for various typesof ordinary differential operators. Techniques and examples which applyto partial differential operators can be found in Friedman [3.4], Stakgold[3.11], Morse and Feshbach [3.8], and Bergman and Schiffer [3.1].

3.3 Inversion of nth-Order Differential Systems

In Section 3.1 we determined the Green’s function and boundary kernelfor a simple second-order system, (3.1). The Green’s function andboundary kernel for the general nth-order differential systems of Section3.2 cannot be determined by the direct integration technique used for thatsimple system, In this section we describe general procedures for solving(3.32) and (3.34) to obtain k and p for the nth-order differential system(3.21) with n independent boundary conditions. The procedures aredemonstrated in detail for regular second-order variable-coefficientdifferential systems.

Obtaining a Complementary Function

Most techniques for determining particular solutions to differentialsystems are based on the complementary function (3.19). Techniques fordetermining the Green’s function k and the boundary kernel p also dependheavily on the complementary function (or the equivalent, a fundamentalset of solutions). In point of fact, the individual segments or components ofk and p are of the form of the complementary function.

I t i s w e l l k n o w n t h a t t h e c o m p l e m e n t a r y f u n c t i o n f o r aconstant-coefficient differential operator consists in sums of exponentials.Let L of (3.18) be the constant-coefficient operator

LiD”+a,D”-‘+.-a +a,,1 (3.36)

To find which exponentials are contained in the complementary functionfor L, we insert a particular exponential v(t) = e@ into the equation Lf = 8

*Integral operators are continuous, whereas differential operators are not. See the discussionof continuous operators in Section 5.4.

Sec. 3.3 Inversion of nth-Order Differential Systems 113

and solve for p. The result is

(3.37)

This equation, known as the characteristic equation for L, has n roots

ppp2,..*,1yt* If the n roots are distinct, the complementary function is

f,(t) = cl exp( pit) + . . * + cn exp( Pnt) (3.38)

Equation (3.38) can be verified by substituting f, into Lf = 9. If two rootsare equal, say, p, = p2, then the corresponding fundamental solutions in(3.38) must be replaced by ci exp( ,uif) + c,t exp( ,uit). This equal root case isdiscussed further in Section 4.4.

We are unable to deal with the variable-coefficient operator (3.18) withmuch generality. An approach that can be used to seek the complementaryfunction for the variable-coefficient operator is the power series method(the method of Frobenius). The method consists in assuming a powerseries form for the complementary function, substituting the series into thehomogeneous differential equation, equating the coefficient on each powerof t to zero, and solving for the coefficients of the power series. The sum ofthe series, where it converges, is at least part of the complementaryfunction. The sum will not, in general, consist of elementary functions. Forexample, Bessel functions arise as fundamental solutions to Bessel’sequation (a second-order variable-coefficient differential equation); thepower series method provides an expression for one of the twofundamental solutions to Bessel’s equation. In the event that the powerseries method does not provide a full set of fundamental solutions for thedifferential equation, other methods must be used to complete thecomplementary function. See Ince [3.6] or Wiley [3.13, p. 255].

Example 1. Power Series Method—Variable Coefficients Suppose

(Lf)(t) : f’(t)+ tf(t) (3.39)

We find the complementarygeneral form

function for (3.39) by assuming a power series of the

fc(t)=t=(Co+CIt+C2t2+ * * *)

where the constant a allows for nonintegerhomogeneous equation and regroup terms:

powers of t. We first insert f, into the

f(t)+ tf(t)=acOta-’ +(a+l)c,ta+[(a+2)c2+c0]ta+l

+[(a+3)c,+c,]ta+2+[(a+4)cq+c2]tu+3+ -es

=o

114

Equating each coefficient to zero, we obtain


We assume, withoutarbitrary; then

loss of generality, that c,+O. It follows that a=0 and co is

and

Determination of the Green’s Function and Boundary Kernel—An Example

We solved for the kernel functions k and p associated with (3.1) by directintegration of the differential equation. Unfortunately, that simpleapproach does not apply to most differential equations. In the followingexample we introduce a general technique for finding k and p.

The model for a particular armature-controlled dc motor and load is thedifferential equation

;i)(‘)+$(t)=u(t) (3.40)

where u(t) is the armature voltage at time t and +(t) is the angular positionof the motor shaft relative to some reference position. Let the boundaryconditions be

+(O)= a, and C/D(~) = a2 (3.41)

That is, we seek the “trajectory” (or angular position versus time), of the


shaft in order that it be in position (xi at time 0 and pass through positiona2 at time b. Comparing this problem to that of (3.21), we note thatL=D2+Q &(+)=W), and #12(+)=+(b). The symbol + replaces thesymbol f used earlier.

Finding the Green’s function for the differential system (3.40)-(3.41) isequivalent to exploring the trajectory $B of the motor shaft for all possibleapplied voltages u(t), but for (pi = a2 = 0. The Green’s function must satisfy(3.32):

Clearly k(t,s) satisfies the homogeneous differential equation in each ofthe regions [O,s) and (s, b]; that is, in the regions where 6 (t - s) is zero. Welet k(t,s)=f,(t) for each of the two regions [O,s) and (s,b]:

k(t,s)=cl+c2e-‘, t in [O,s)

=dl+d2e-‘, t in (s,b]

Since k(t,s) is a function of s, the arbitrary constants must depend on s.We eliminate half of the arbitrary constants by applying the boundaryconditions

k(O,s)=c,+c,=O + c2= -cl

k(b,s)=d,+d2eBb=0 + d2= -ebdl

It is the second (or highest) derivative of k that introduces the deltafunction in (3.32); for if the first derivative included a delta function, thesecond derivative would introduce the derivative of the delta function.*Since d2k/dt2 includes a unit impulse at t = s, dk/dt must include a unitstep at t = s, and k itself must be continuous at t = s. We express these factsby the two “discontinuity” conditions:

*See Appendix 2 for a discussion of unit steps, delta functions, and derivatives of deltafunctions.

116

Applying these conditions to k(t,s), we find


-d2e-S-(-c2e-J)= 1

A messy elimination procedure among the boundary condition equationsand discontinuity condition equations yields

cl(s)= e9--eb and d,(s) = e”-leb- 1 eb-1

It follows that

k(t,s)=(1 - e-‘)(e’- e”)

eb-1?Qs

(3.42)

(1 - ebe-‘)(e’- 1)=

eb-1t>s

To get a feel for the nature of this system (for which +(O)=+(b) = 0), weuse k to determine the shaft trajectory Q, and velocity profile C$ for aspecific input u(t) = 1:

+(t)= lbk(t,s)u(s)h0

= l-ebe-’ *eb_ l I, (es- l)dr+ yJb(eS-eb)dS

-1 c

= t-( 1--j$+ (l-e-‘)

+(t)=l- -j$ e-’( 1

The trajectory $D and the velocity profile i are plotted in Figure 3.2 forb = 1. Observe that, in general, the motor shaft cannot be at rest at t = 0and at t = b if the shaft positions are specified; it is precisely the freedomin the initial and terminal velocities which allows us to choose both the endpoints, #B(O) and +(b), and an arbitrary continuous input voltage u.

The boundary kernel p for the system (3.40)-(3.41) describes the tra-jectory rg(t) as a function of the boundary conditions HO)= (ri and


Figure 3.2. Shaft position and velocity for +(O) = #(l) = 0 and u(t) = 1.

C/B(~) = LYE with no voltage applied to the motor; that is,

Perhaps the most direct approach to the determination of pi is to let+(t) = crv,( I) + c2v2( t), a linear combination of the fundamental solutionsfor (3.40), then apply the boundary conditions (3.41) to obtain thecoefficients Ci as a function of CX, and 1~~. Rather than use this approach, weattack the defining equations for pi in a more formal manner whichparallels the determination of the Green’s function. The two approachesare equivalent in the amount of computation they require. The boundarykernel satisfies (3.34):

iqt>+bl(t)=o i&(t) +b2(t) =o&(p,)=p,(O)= 1 &(P2)=Pz(O)=O

JB,(p,)=p,@)=O JB2(P2)=PZw= l

The boundary condition statements are reminiscent of the boundarycondition matrix (3.27). In point of fact, pt and p2 each consist in a linearcombination of the fundamental solutions v,(t) = 1 and v2( t) = e-t. Apply-


ing the boundary conditions to p,(t) = c, + c2e -‘, we get

or

where B is, indeed, the boundary condition matrix of (3.27). Similarly,using p2( t) = d, + d2e - ‘, we find

We can combine the two coefficient equations into the single matrixequation

which has the solution

The function pj is a specific linear combination of the two fundamentalsolutions specified above; the jth column of B-’ specifies the linearcombination. Thus

ebm= -& + -p-y

p2(t)= L + LIZLesteb-1 eb-1

(3.43)

The shaft position and velocity, as functions of the boundary conditions,are


Figure 3.3 shows the position and velocity of the motor shaft for ai = 0 andIY~ = 1. The shaft is already in motion at t = 0, and exhibits an “undriven”decay in velocity until it reaches the position +(b) = 1 rad. If the boundaryconditions were ai = a2 = 1, the shaft would sit at rest in the position+((t) = 1 rad; again an undriven trajectory.

The inverse of the system (3.40)-(3.41) is the sum of the separatesolutions for the distributed and boundary inputs. That is,

where k and p are given in (3.42) and (3.43), respectively. The nature of thesystem (3.40)-(3.41) does not seem in keeping with the nature of dynamic[real-time) systems. The motor must anticipate the input u(t) (or theImpulse 6 (t - s)) and appropriately select its velocity at t = 0 in order to beable to meet the requirement on its position at t = b. We are more likely tomeet such a two-point boundary value problem when the independentvariable t represents not time, but rather a space variable. Yet a two-point

90)

I

b

Figure 3.3. Undriven shaft position and velocity for a, = 0 and a2 = 1.


boundary value problem can arise in a dynamic system if we imposerequirements on the future behavior of the system as we did in (3.41).

Summary of the Technique

The technique demonstrated above for determining the Green’s functionand the boundary kernel depends upon knowledge of the complementaryfunction. We can apply the technique to the regular nth-order system(3.21) if the corresponding complementary function can be determined.Assume L of (3.21) has the complementary function f = crvr + . . . + c,v,.Further assume that the system is invertible (i.e., we have n independentboundary conditions for which (3.28) is satisfied). We obtain the Green’sfunction k and the boundary kernel p for the system (3.21) by followingthe technique used for the system (3.40)-(3.41).

Equation (3.32) determines the Green’s function k. The unit impulse6 (t - S) is zero for all t # S. Therefore, k(t,s) satisfies the homogeneousdifferential equation for t#s; k(t,s) is equal to the complementary func-tion (3.19) in each of the two regions [a,~) and (s, b]. Because the comple-mentary function f, is used in two separate regions, we must determine twosets of n arbitrary constants:

k(t,s)= blv,(t)+ . +. + b,,v,,(t), t in [a,s)= d,vl(t) + . m . + dnvn(t), t in (s,b] (3.44)

Half of the 2n constants can be eliminated by the homogeneous boundaryconditions of (3.21): &k(t,s)=O, i=l , . . . ,n. The rest are determined byappropriate “discontinuity” conditions at t = S. Only the highest derivativeterm, g,(t)d”k(t,s)/dt”, can introduce the delta function into (3.32)(otherwise derivatives of delta functions would appear); therefore, wematch the two halves of k(t,s) at t= s in such a way that we satisfy thefollowing n conditions:

kdk’ dt

)..., dn-2kdtn-2

are continuous at t = s

d”-‘k(s+,s) d”-‘k(s-,s) 1#I-’ -

=-dt”-’ go@)

(3.45)

That is, d”-‘k(t,s)/dt”-’ must contain a step of size l/g,(s) at t=s. Theng,(t)d’k(t,s)/dt” will include the term 8 (t-s).*

*See Appendix 2 for a discussion of steps, delta functions, and derivatives of delta functions.


The boundary kernel p is specified by (3.34). Each component of p is alinear combination of the fundamental solutions for L:

pj=cljvl+--- +cnjvn j=l,...,n (3.46)

Applying the n boundary conditions of (3.21) as required by (3.34), we find

St(Pj). =

i H

Sr(vl) ’ ’ ’ Bl(‘n). .. . .

Pn iP,> Pn iv,> � l � i (�n)

0clj

...

)Ol i

=. ij , j=l,...,n. .

c . ln/

d

These n sets of equations can be expressed as

(3.47)

It follows that the coefficients for pj in (3.46) are the elements in the jthcolumn of B-l, where B is the boundary condition matrix defined in(3.27). Specifically, cO is the element in row i and column j of B-l.

Exercise 1. Let f’(t)+ tf(t)=u(t) with f(O)= (pi. (The complementaryfunction for this differential equation was determined in Example 1.) Showthat the inverse of this differential system is

f(t) = exp( - t2/2)itexp(s2/2)u(s) ds + aI exp( - t2/2> (3.48)

Second-Order Differential Systems

Many of the ordinary and partial differential equations that arise in themodeling of physical systems are second order. Some of the second-orderpartial differential equations can be reduced, by a substitution of variablesor by integral transforms, to second-order ordinary differential equations.*Furthermore, use of the “separation of variables” technique in solvingsecond-order partial differential equations produces sets of second-order

*See Kaplan [3.7].


ordinary differential equations. Thus the general second-order ordinarydifferential equation with variable coefficients is of considerable practicalimportance. We present explicit expressions for the Green’s function andthe boundary kernel for an arbitrary regular second-order differentialsystem; these expressions are obtained in terms of a fundamental set ofsolutions for the differential operator.

The regular second-order differential system is†

go(t)r”(t) + g,(t)f’(t) + g2(9f(t) =w

km = a1 and P2(f) = (r2(3.49)

where gi is continuous and g,(t)#O in the region of interest. Assume vr andv2 are independent solutions to the homogeneous differential equation. By(3.44), the Green’s function is of the form

k(t,s)= blv,(t)+ b2v2(t), t<s

= d,v,( t) + d2v2( t), t > s

The discontinuity conditions (3.45) become

Since dk/dt has a step of size l/g,(s), then go(t)d2k/dt2 includes a unitimpulse. These two discontinuity equations can be put in the matrix form

The solution is

d,- b,= -v2m

d2- b,=Vl (4

w(s) go(s) ’ w(s) go(s)

where w(s) is the Wronskian determinant*:

(3.50)

† In Section 5.5 we refer to the differential operator of (3.49) as a regular Sturm-Liouvilleoperator.*Note that the solution is undefined for w(s) = 0. It can be shown that if v1 and v2 areindependent solutions to the homogeneous differential equation, then w(s)#O for all s in theinterval of interest. See P&C 3.7.


The boundary conditions &k (t, s) = P2k( t, s) = 0 provide two more linearalgebraic equations which, together with the above pair of equations,determine the constants b,, b,, d,, and d2, and therefore, k(t,s). However,without specific information about the nature of the boundary conditions,we can carry the solution no further. The solution for a dynamic system(initial conditions) is given in Exercise 2. Two-point boundary conditionsare treated in Exercise 3.

Exercise 2. Let the boundary conditions of (3.49) be

p*(f) 2 f(a) a n d #12(f) i f’(a) (3.5 1)

Show that the corresponding Green’s function is

k(t,s)=O, t in [a,s)

A@, t) (3.52)= go(s)w(s) 9 t in (s, 4

where w is given by (3.50), and

Show also that the corresponding boundary kernel is

PdO =v;(4w) - 6 m2w

44(3.53)

P2W =vd4v2W -v2Wvdt)

w(a)

Exercise 3. Let the boundary conditions of (3.49) be fir(f) 4 f(a) and

f12(f) i f(b). Show that for this two-point boundary value problem

k(t,s)= ’goww

rA(b,s)A(a, t)

A(a,b) ’a<tQs

A(b,s)A(a,t)

A@, b)+ A@, t), s<t<b

A(& t)PI(t)= - -

A@, t)

A@, b) and P2(9= qgq

(3.54)

where A(s, t) is given beneath (3.52) and w(s) is defined in (3.50).


Exercise 4. Use (3.54) to find k and p for the dc motor system (3.40)-(3.41). Compare the result with (3.42) and (3.43).

It is apparent that we could derive an explicit expression for the inverseof a regular nth-order linear differential system [assuming the boundaryconditions satisfy the invertibility condition (3.28)]. The inverse wouldinvolve n independent fundamental solutions and the nth-order Wronskiandeterminant of these n solutions. Of course, as indicated by Exercise 3, themanipulation can be complicated. The determination of the Green’s func-t ion for an nth-order two-point boundary value problem requires thesolution of 2n simultaneous algebraic equations with coefficients which arefunctions of s. In contrast, the Green’s function for the initial conditionproblem (or one-point boundary value problem) requires the solution ofonly n simultaneous equations because k( t,s) = 0 for t < s. Of particularinterest is the constant-coefficient initial condition problem, for whichdetermination of the Green’s function reduces to inversion of an n x nmatrix of constants.

3.4 Time-Invariant Dynamic Systems

The initial value problem is at the heart of dynamic systems-systems forwhich the variable t represents time. The linear time-invariant (orconstant-coefficient) dynamic system merits special attention if onlybecause its inversion is easily automated using standard computerprograms for solving matrix equations. Furthermore, many dynamicsystems are adequately represented as linear time-invariant systems. Weexamine these systems in detail in this section.

The Inverse of the nth-Order System

The general nth-orderconditions is

constant-coefficient differential equation with initial

P(t)+a,r(“-l)(t)+ - -. +a,f(t)=u(t)

P,(f) % f(i-l)(0)=ai i = l,...,n

(3.55)

for real scalars {a,.} and t 2 0. The characteristic equation for (3.55) is(3.37); assume it has n distinct roots pl,. . . ,P~ (the multiple root case isconsidered in Section 4.4). Then the fundamental solutions for (3.55) are

vi(t) 2 exp( pit), i= l,..., n.

Sec. 3.4 Time-Invariant Dynamic Systems 125

The Green’s function, as given by (3.44), is

k(t,s)=b,exp(p,t)+ *a* +b,exp(kt), tin [O,s)

=d,exp(p,t)+... + dn exp( pnt), t in (s, 00)

All n boundary conditions apply to the first half of k( t, s), the halfinvolving the unknowns b,, . . . , b,. As a result,

This boundary condition matrix is the Wronskian matrix of the functions{ exp( hit)} at t = 0. The matrix is also known as the Vandermond matrix forthe system (3.55). It is invertible if and only if the roots pl,. . . ,pn aredistinct as assumed.* Therefore, b, = . . . = b,, = 0, and k(t,s) = 0 for t in[O,s). The discontinuity conditions (3.45) at t = s are

d, exp( pts) + . . . + d,exp( fins)=0 (k continuous)

d, 1-11 exF$ PI’> + * * ’ + dnPn exP( l-$zs) = O (dk/ dt continuous). .

4 kc2 exp’( pg) + . . - + d,&+‘exlj( hs) = 0 ( dnS2k/dt n-2 continuous)

d,y;-’ exp( pts) + . . . + d,&‘- ‘exp( pns) = 1 (unit step in d”-‘k/dt”-‘)

We substitute the new variables 4 2 di exp( piis), i = 1,. . . , n into the discon-tinuity equations to obtain

(3.56)

*If the roots were not distinct, we would use a different set of fundamental solutions {vi}, andobtain a different boundary condition matrix. The Wronskian matrix is explored in P&C 3.7.The Vandermond matrix is examined in P&C 4.16.


Because the roots { pi} are distinct, the Vandermond matrix is invertible,and (3.56) can be solved by means of a standard computer program toobtain {di}. Notice that the new variables { di} are independent of s. The sdependence of the variables { 4} has been removed by the substitution. Interms of the new variables, the Green’s function becomes

k(t,s) =0 for 0 < t < s=d,exp[ pl(t-s)]+. . * +$exp[ pn(t-s)] fort 2 s

(3.57)

The boundary kernel for the system (3.55) is found from (3.46) and(3.47). Equation (3.47) is

Then, by (3.46),

pi(t)=cvexp(lFllt)+... +cnjexp(pnt), j=l,...,n (3.58)

where the coefficients for pi are obtained from the jth column of theinverse Vandermond matrix:

The inverse of the differential equation and boundary conditions of(3.55) is

(3.60)


where {4} and { cV} are specified by (3.56) and (3.59), respectively. Thecomputer program which produces (3.59) will simultaneously solve (3.56).Section 4.4 explores the computational difficulties which arise when thecharacteristic equation of the system has nearly equal roots.

The shape of the Green’s function for a time-invariant (i.e., constant-coefficient) dynamic system depends only on t-s, the delay between thetime s that an impulse is applied at the system input and the time t that theoutput k(t,s) is observed. That is, k(t,s)= k(t - s,O). Therefore, actualmeasurement of the response of the physical system to an approximateimpulse is a suitable method for determining the Green’s function. Theresponse of such a system, initially at rest, to an impulse input u(t) = S (t) iscommonly referred to as the impulse response of the system. We denote the

impulse response by g, where g(t) 4 k(t,O). Then the integral term in(3.60) can be rewritten as a convolution of u and g.*

[‘k(t,s)u(s)ds= Stg(t -s)u(s)ds0 0

The components of the boundary kernel also can be measured physi-cally; pj(t) is the response of the system with no distributed input u, andwith the initial conditions CXJ = 1, Cwi(3.56)-(3.59) that L&= tin for i= 1

= 0, for i#j. Furthermore, we see from, . ..,n. Therefore, the impulse response is

equal to one of the initial condition responses; specifically,

Pn(t)= k(t,O)=g(t) (3.61)

Applying a unit impulse 8(t) is equivalent to instantaneously applying tothe system (at rest) the unit initial condition f@- “(0) = 1 (all other initialconditions remaining zero); if we can apply this initial condition someother way, we do not need an approximate impulse in order to measure theimpulse response of the system.

Exercise 1. The differential equation (3.40) for an armature-controlled dcmotor is

Show that for given initial conditions, +(O) and C&O), the Green’s function,

*See Appendix 2 for a discussion of convolution.

128

boundary kernel, and inverse equation are


k(t,s)=O, O<t<s= l_ ,-(t-s), tas

PI(t) = 1

p2(t)=l-e-’(3.62)

+(t)=jo’[l-e-“-“)]u(S)dS++(O)+(l-e-$(O)

Compare (3.62) with (3.52) and (3.53).

The State-Space Model

The nth-order constant-coefficient differential equation with initial condi-tions, (3.59, can be expressed as a first-order vector differential equationby redefining the variables. If u(t) = 0, the quantities f(O), f(‘)(O), . . . , fi”- “(0)determine the trajectory f(t) for all t; these n quantities together form amore complete description of the state (or condition) of the system at t = 0

than does f(0) alone. Let f, % f, f, e r”), . . . ,fn A r<“- l). Then (3.55) can beexpressed as the following set of n first-order differential equations.

I, =f2W

i,(t) = f3W

i,- l(t) =mwi,(t) = - a,f,( t) - . l l - dlw + u(t)

By defining x 3 (f, . . . f,JT and i b (ii . l l ill)=, we write the n individualequations as

(3.63)

The square matrix of (3.63) is known as the companion matrix for the n th


order differential operator (3.55). The initial conditions of (3.55) become

(3.64)

We call x(t) the state vector of the system at time t. Since the differentialsystem (3.55) has a unique solution, the state at time t can be determinedfrom the state at any time previous to t. The state provides preciselyenough information concerning the condition of the system to determinethe future behavior of the system for a given input. The vector x(t) is in9lLn x ‘. Therefore, we call ‘%,“’ x ’ the state space of the system. Thevariables {fi(t)} are known as state variables.

Equations (3.63) and (3.64) are of the general form

k(t)=Ax(t)+Bu(t), x(0) given (3.65)

However, the notation of (3.65) is more general than that of (3.63) and(3.64). The input u can include more than one function. A meaningfulequation is defined by any n x n matrix A and nX m matrix B; theresulting vector equation describes the evolution in time of a system withm inputs and n outputs. A general set of coupled linear time-invariantdifferential equations can be expressed in this state-space form (P&C3.18). We refer to (3.65) as a state equation. We call x(t) the state vectorand its elements the state variables; A and B are the system matrix and theinput matrix, respectively.*

We should note that the description of a dynamic system by astate-space model is not unique. If we multiply both sides of (3.65) by anarbitrary invertible n x n matrix S, we obtain

Srit( t) = SAx( t) + SBu( t)

Defining y = Sx, we find

f(t) = SAS- ‘y(t) + SBu( t)

=Ay(t)+h(t)

*See Zadeh and Desoer [3.14]discussion of state-space models.

or DeRusso, Roy, and Close [3.2] for a more complete


with y(O)= Sx(0) given. This second state-space differential equation isequivalent to (3.65) as a representative of the system. The state vector y(t)is a representation of x(t) in new coordinates. Thus the state variables andsystem matrix which describe a given system are not unique. In Section 4.2we explore the essential characteristics of a matrix, its eigenvalues. We findthat the similarity transformation SAS-’ does not affect the eigenvalues.Consequently, all system matrices which represent the same system havethe same essential characteristics. State space models of dynamic systemsare analyzed in terms of their eigenvalues in Sections 4.3 and 4.5.

Example 1. A State Equation. The differential equation for the armature-controlled dc motor of Exercise 1 is

m +&t> =u(t), &u = aI, MO = (3

Defining the state variables f,(t) i #(t) and f2(t) 2 c&t), we obtain the followingstate equation

*w=(; Jxw+( !+(I), x(0)=( Z;)

The systemequation.

matrix is the companion matrix for the second-order differential

Let us find an integral equation which is the inverse (or explicit solution)of the first-order vector-valued differential system (3.65). Although wework directly with the system in the specific form (3.65), we note that theequation can be expressed in terms of a general differential operator Lacting on a vector-valued function space. Let f be in (9 (0, co); then fik) isin and x is in the Cartesian product space:

?r= eye, co) x P-‘(0, 00) x * * - x eye, co)

The system (3.65) is equivalent to the following operator equation on ‘V:

LX b jL -Ax = Bu with x(O) given (3.66)

We express x as an integral operation on the whole vector-valued functionBu.

Inversion of the State Equation

The state equation for an nth-order time-invariant dynamic system is

k(t)=Ax(t)+Bu(t), x(O) = x0 (3.67)


where A and B are arbitrary n X n and n X m matrices, respectively. Thestate vector x(t) and the input vector u(t) are in wx ’ and TILmx ‘,respectively. We invert (3.67) by the same approach we used for thenth-order differential system; we invert separately the two componentequations

k(t)-Ax(t)=Bu(t), x(0) = 9 (3.68)

i(t)-Ax(t)=& x(O) = x0 (3.69)

Assume the inverse of the “boundary input” system (3.69) is of the form

x(t) = q t)x(O) (3.70)

where the boundary kernel @(t) is a n X n matrix commonly referred to asthe state transition matrix. (It describes the “undriven” transition from thestate at “0” to the state at t.) In order that (3.70) be the correct inverse, x(t)must satisfy (3.69),

for any initial condition vector x(O). Therefore, the state transition matrixmust satisfy

(3.7 1)

Rather than treat the system (3.71) one element at a time, we work withthe whole n x n matrix-valued system. We use the power series method tofind the complementary function for the system. Assume

where each Ci is a constant n X n matrix. We substitute @(t) into thedifferential equation of (3.71) and equate the coefficient on each power of tto the zero matrix 8 to find

It follows that Co is arbitrary and

(3.72)


We have used the symbol eAt to represent the sum of the “exponential-looking” matrix series of (3.72):

A2t2e”lAI+At+-?;-a--

.

We call eAt a fundamental matrix for the state equation of (3.67); thematrix is analogous to a fundamental set of solutions for an nth-orderdifferential equation. Applying the boundary conditions of (3.71) to (3.72),we find m(O) = eAoCo = I. It is clear from the definition of eAt that eAo= I;therefore, Co = I and the state transition matrix (or boundary kernel) forthe state-space system (3.67) is

*p(t) = eAt (3.73)

Example 2. A State Transition Matrix. In Example 1 we found the systemmatrix A for the differential equation &t) + &t) = u(t):

To find the fundamental matrix for this system, we sum the defining infinite series:

If the matrix A of Example 2 were not simple, it would be difficult tosum the infinite series for eAr by the method of that example. It would notbe easy to recognize the function to which each scalar series converges.Arbitrary functions of matrices are examined in detail in Section 4.6, andpractical techniques for computing functions of matrices are developed.These techniques can be used to compute eAt for an arbitrary squarematrix A.


Exercise 2. Show that for the fundamental matrix eAr of Example 2,

(3.74)

Properties (3.74) apply to all time-invariant systems, that is, all systemswhich have a constant system matrix (P&C 3.19).

We can view Bu as a vector-valued distributed input to (3.67). Therefore,we assume the inverse of the distributed-input state equation (3.68) is anintegral equation of the form

x(t)= imK(t,s)Bu(s)dr (3.75)

where the n X n matrix K(t,s) is called the matrix Green’s function for thesystem (3.67). (By the integral of a matrix we mean the matrix of integrals.)We substitute (3.75) into (3.68) to determine the equations which describeK:

(3.76)

for all vectors u with elements which are continuous functions. To seemore clearly the conditions on K(t,s) which follow from (3.76), note that

In other words, if we let G(s) denote the matrix with elements g#(s), thenthe equation /FK( t, s)G(s) dr = G(t) is satisfied by K(t,s) = 6 (t - s)I. Thusin order to satisfy (3.76), it is sufficient that K meet the followingrequirements:

dK(t,s)F -AK(t,s)=6(t-s)I

dtK(O,s) = 9

(3.77)


The approach we use to solve (3.77) for K is essentially the same as thatused for the nth-order scalar system (3.55). For t#s, K(t,s) satisfies thesame n x n differential equation, (3.71), as does the state transition matrix.Thus, using the general solution to (3.71) found earlier,

K( t, s) = eAtBO, t in [O,s)

= e*‘D0, t in (s, 00)

where B. and Do are n x n constant matrices. The boundary conditions of(3.76) require K(0, s) = eAeBo = 8 ; since eAo = I, B, =8. From (3.77), we alsonote that K must satisfy a discontinuity condition at t =s. The deltafunctions on the right-hand side of (3.77) must be introduced by thehighest derivative, dK/dt; otherwise derivatives of delta functions would appear. Consequently, the diagonal elements of K contain a unit step att = s, whereas off-diagonal elements are continuous:

K(s+ ,s)-K(s-,s)=ehDO-8-I

Then, using (3.74), Do= (e”)- ’ = ehkr, and

K(t,s) =8, t<s= e*bs) , t>s

(3.78)

The- inverse of the state-space system (3.67) is the sum of (3.75) and(3.70); Q, and K are given by (3.73) and (3.78), respectively:

(3.79)

The inverse system is fully determined by the state transition matrix eAtand the input matrix B. In Section 4.6 we determine how to evaluate eAt bymethods other than summing of the series (3.72).

At the heart of the solution (3.60) for the nth-order dynamic system(3.55) is the Vandermond matrix for the system. If the state equation isderived from the nth-order differential equation as in (3.63), we wouldexpect the Vandermond matrix to be involved in the solution (3.79) of thestate equation. We find in P&C 4.16 and (4.98) that if the system matrix Ais the companion matrix for an nth-order dynamic system, the Vander-mond matrix is intimately related to both A and eAt.

Exercise 3. Show that for the system of Examples 1 and 2,

x(t)= ($) = I’( ’ $;‘))u(s)ds+ ( *(“)+;;$;;-t’o)e-t) (3.80)

Sec. 3.5 Problems and Comments 135

Equation (3.80) should be compared with its second-order scalar equiva-lent (3.62). The state-space solution usually contains more informationthan its scalar counterpart-information about derivatives of the solutionis stated explicitly.

Exercise 4. Use the solution (3.79) at t = a to determine the form of thesolution to the state-space system (3.67) if the initial conditions are given att = a instead of t =O; that is, show that

f(t)= p?A(t-S)Bu(s)ds+ eA(‘-u)x(a)a

The discussion beneath the nth-order scalar solution (3.60) extends tothe more general state-space solution (3.79). We can interpret K(t, 0) = eAras the matrix impulse response of the state-space system. Since the matrixA is constant, it is appropriate to measure physically the state transitionmatrix eAt. By (3.70), the jth column of a(t) (or e*‘) consists in the“undriven” decay of x(t) from the initial condition x(O)= 5, the j t hstandard basis vector for %Y x ‘. From measurements of the n columns ofeAt we can determine the full inverse equation (3.79) without explicitdetermination of the system matrix A (P&C 3.20).

The techniques used to invert the first-order state-space system (3.67) areapplied to a second-order vector differential system in P&C 4.32. As withthe state-space system, the Green’s function for this system can beobtained from the boundary kernel; the latter can be measured physically.The inverse for this second-order vector system involves several functionsof matrices. We discuss methods for evaluating general functions ofmatrices in Section 4.6.

3.5 Problems and Comments

3.1 Forward integration: the differential system f”(t) + ;f( t) + (l/400)f3(t)= 0, f(0) = 10, f’(O)= 0 describes the unforced oscillations of amass hanging on a spring. The spring has a nonlinear force-elongation characteristic; f(t) denotes the position of the mass attime t. There are many numerical integration techniques for obtain-ing an approximate solution to such a nonlinear differential equa-tion with initial conditions (see [3.9]). The following technique isone of the simplest. We concern ourselves only with integer valuesof t, and replace the derivatives by the finite-difference approxima-tions f’(n)mf(n+ 1)-f(n) and f”(n)mf(n+ l)-2f(n)+f(n- 1). Usethese finite-difference approximations and the differential system


to express f(n + 1) in terms of f(n) and f(n - 1). Computef(M%.., f(8). How might the above finite-difference approxima-tion be modified to obtain a more accurate solution to the differen-tial equation?

3.2 Backward integration: a (nonlinear) differential equation with finalend-point conditions (rather than initial conditions) can be solvedby backward numerical integration. Backward integration can becarried out be means of any forward integration routine. Supposethe d i f fe rent ia l sys tem i s o f t h e f o r m f(“)(t) + F(f( t),

w ,..., f@-‘)(t),t)=O with f(t,>,f’(t,> ,..., r’“-‘I($) specified. Showthat the change of variables f(t) = f(+- s) = g(s) converts the finalconditions on f to initial conditions on g and produces a differen-tial equation in g which differs from the differential equation in f inthe sign on the odd-order derivatives.

3.3 Relaxation: the finite-difference approximation to a two-pointboundary value problem can be solved by a simple iterativetechnique known as relaxation [3.3]. Suppose f”(s)= 1 with f(0)= f(5)=0. Consider the values of f only at integer values of s.Replace the second derivative by the approximation f”(n)af(n +1) - 2f(n) + f(n - l), and express f(n) in terms of f(n - 1) and f(n +1). Let the initial values of f(l), . . . , f(4) be zero. A single step in theiteration consists in solving successively for each of the values

f(l) , . . . , f(4) in terms of current values of f at the two neighboringpoints. Repetitive improvement of the set of values {fm(k)} results inconvergence of this set of values to the solution of the set ofdifference equations, regardless of numerical errors, and regardlessof the order in which the values are improved during each iteration.(a) Carry out six iterations for the above problem.(b) Find the exact solution to the set of difference equations by

solving the equations simultaneously. Compare the results ofthe iteration of (a) with the exact solution for the differentialsystem.

*3.4 An intuitive understanding of the following properties of differen-tial systems can be gained by examining a finite-differenceapproximation to the second-order case. See [3.6] for a rigorousdiscussion of these statements.(a) A regular nth-order linear differential equation has n inde-

pendent solutions.(b) A boundary condition consisting in a linear combination of

values of f, f’, . . . , fin-l) need not be independent of the regularn th-order differential equation; consider, for example, f”(s)= 0 with f’(0) - f’( 1) = 0.


(c) If the boundary conditions associated with a regular nth-orderdifferential equation consist in n independent linear combina-tions of the values f(a), f’(a), . . . , P - “(a), at a single point a inthe domain of f, then the differential system has a uniquesolution.

3.5 The following differential system is degenerate:

Find the solutions to the differential system in terms of the inputsu, ai, a2, and (x3. Also find the relations among the inputs that mustbe satisfied in order that solutions exist. (Hint: the solution to thedifferential equation is expressed in terms of +(O) and #(O) in(3.80).) What relationship exists between the number of dependentrows in a boundary condition matrix for a system and the numberof different relations which must be satisfied by the inputs to thatsystem?

3.6 Let L be a regular nth-order differential operator and { &.(f) = 0,i = 1 , . . . , m} a set of homogeneous boundary conditions. Let V bethe space of functions in P (a, b) which satisfy the homogeneous

differential equation Lf = 8. Let 9 i {vi, v2,. . . ,vn} be a funda-mental set of solutions for L; 9 is a basis for V. Define T:

Ir+w by Tf 9 (&(f),...,/3m(f’)) for all f in V. Let & be thestandard basis for CR,“. Show that the matrix [T],, is a boundarycondition matrix for the differential system {L,&, . . . ,&,}.

*3.7 The Wronskian: let f,, . . . , f, be in C? (a, b). The Wronskian matrixof fi, . . . , f, at t is defined by

The Wronskian determinant is w(t) i det(W( t)).(a) S h o w t h a t {fi,...,f,} cannot be linearly dependent unless

w(t)=0 for all t in [a,b].


(b) The fact that w(t) = 0 for some t does not ordinarily imply thatthe set {f,, . . . , f,} is dependent; try, for example, f r( t) = t2 andf2(t) = t3 at t = 0. Suppose, however, that f,, . . . ,f, are solutionsto an nth-order homogeneous differential equation defined on[a, b]. Then if w(t) = 0 for any t in [a, b], {f,, . . . , f,} is a linearlydependent set.

3.8 Difference equations: an arbitrary linear constant-coefficientdifference equation can be expressed in the form

a,(E”f)(k) + a,(E*- ‘f)(k) + - - * + a,,f(k) =u(k), k=O, 1,2,...

where E is the shift operator defined by (Ef)(k) 9 f(k + 1); weconcern ourselves only with integer values of the argument of f.The order of the difference equation is the number of boundaryconditions needed to specify a unique solution to the equation; thatis, the order is n -p, where p is the lowest power of E to appear inthe equation. (See [3.2].)(a) The solutions to the homogeneous difference equation (the

equation with u(k) = 0) usually consist of combinations ofgeometric sequences. Substitution of the sequence f(k) = r k,k=0,1,2 ,..., into the homogeneous equation shows that non-trivial sequences must satisfy the following characteristicequation: aOr’ + a,?-‘+ . . . + an = 0. Find a basis for thenullspace of the difference operator T defined by

(Tf)(k) i 2(E2f)(k)-3(Ef)(k)+f(k)

=2f(k+2)-3f(k+ l)+f(k)

What is the dimension of the nullspace of an nth-orderdifference operator?

(b) Let fl,..., f, be infinite sequences of the form fj(k), k=0,1,2 ,.... The Casorati matrix of f,, . . . , f, is defined by

The infinite sequences f I,. . . , f, are linearly independent if and


3.9

3.10

3.11

3.12

3.13

3.14

*3.15

only if c(k) # 0 for k = 0, 1,2,. . . , where c(k) is the Casoratideterminant, det(C(k)). Use the Casorati determinant to showthe independence of the basis vectors found in (a).

Use the power series method to find the complementary functionfor the differential operator (D- 1)2.

Define L: e’(O, l)+ e (0,l) by L i -D - a1. Find the Green’sfunction k and the inverse equation for the differential systemLf=u, f(O)=f(l).

Define L: e2(0, b)+ (? (0, b) by L i D2- 3D+21. Find the Green’sfunction k, the boundary kernel p, and the inverse equation for thedifferential system Lf = u, f(0) = (x1, f(b) = (r2.

Find the inverse equation for each of the following differentialsystems:(a) f”+6f’+5f=u, f(O)=a,, f’(O)=a,(b) f” + 2f’ + 2f = u, f(0) = a,, f’(0) = a2(c) f”’ + 6f” + 5f’ = u, f(0) = a,, f’(0) = a2, f”(0) = a3The following differential system describes the steady-state tem-perature distribution along an insulated bar of length b: - f” =u,f(O)= a,, f’(b)+f(b)= (Ye. (The second boundary condition impliesthat heat is removed by convection at point b.) Show that theinverse equation for this system is

For the differential system d’(t) -f(t) = u( t), f’( tl) = a, t, > 0,(a) Find the complementary function by the power seriesmethod;(b) Find the Green’s function k( t, s);(c) Find the boundary kernel pi(t);(d) State explicitly the inverse equation.Let pi and p2 be the roots of the characteristic equation for thedifferential system f” + a$’ + a,f = u, f(0) = f’(0) = 0.(a) Use (3.56) and (3.57) to find the Green’s function k for this

system. If p2*pl, computed values of p2- ,ui and exp( p2t)-exp( pit) will be badly in error. What is the effect of nearequality of the roots on the numerical computation of k(t,s)and lk(t,s)u(s)di?


(b) If p2wp1, the fundamental set {exp( p, t),exp( p2t)} is nearlydependent. A better fundamental set (not nearly dependent)in this circumstance is

Derive a power series expansion of v2 which can be used tocompute values of v2 without numerical division by the in-accurate quantity p2 - pi. Show that as p2--+pl, {vl(f),v2(t)}+{exp( l%t)JexP( l%N*

(c) Equation (3.52) expresses the Green’s function k in terms ofthe functions {vi} of (b). Evaluate the Wronskian determinantw in this expression in terms of exponentials. Values of k(t,s)and J k( t, s)u(s) ds can be computed accurately by using thisexpression for k(t,s) together with computed values of vl,v2,and w. Show that this expression for k(t,s) is a rearrangementof the expression for k( t, s) found in (a).

3.16 One method for obtaining the Green’s function for a constant-coefficient differential system is to solve (3.32) by means of one-sided Laplace transforms. Use this technique to show that theinverse of the differential equation i+02f =u, with constant w andgiven values of f(0) and i(O), is

3.17 The approximation of derivatives by finite-differences leads to theapproximate representation of differential equations by differenceequations. For instance, the use of a second-central difference plusa forward difference converts the second-order differential systemc/b” +rp’=u, $3(O)= CY,, +‘(O)= a2 to the approximately equivalentsecond-order difference system 2+( i + 2) - 3+( i + 1) + cp(i) = u( i + I),HO)= ai, +(l)= cr,+ (or. A general form for the n th-order constant-coefficient difference system with initial conditions is

f(i+n)+atf(i+n-l)+.s* +a,f(i)=v(i)

f(O)=u,, f(l)=u, . . . . f(n- l)=y,

for i=O,1,2 ,... .By analogy to the inverse equation for the nth-order differentialsystem, we assume the inverse of the nth-order difference system is

Sec. 3.5 Problems and Comments

of the form

f(i)= 5 k(i,j)v(j> + 5 p,(i)f(m- 1)j-0 m=l

141

for i=O,1,2 ,... .(a) Show that the discrete Green’s function k(i,j) is specified by

the difference system

for i=O,1,2 ,... and j=O, I,2 ,... .(b) Show that the discrete boundary kernel p,(i) is specified by

the difference system

p,(i+n)+a,p,(i+n- I)+ a** +anp,(i)=O

PAP) = s,,+ 1

for i=O,1,2 ,..., m= l,..., n, and ,..., n- 1.(c) Find the inverse of the second-order difference system

mentioned above by solving the difference systems corres-ponding to those in (a) and (b). Hint: solutions to homo-geneous constant-coefficient difference equations consist insums of geometric sequences of the form f(i) = r i, i = 0, + 1, &2,... .

3.18 The following pair of coupled differential equations relates a pairof system outputs {fi(t)} to a pair of inputs {U,(t)} :

f;’ +3f;+2f2=u,,

f; +r;+r,=u,,

f1(0),f;(0),f2(O),f;(O) specified.

(a) Find a first-order state equation of the form (3.65) which isequivalent to the set of coupled equations. (Hint: use as statevariables the output functions and their first derivatives.) Isthe state equation unique?

(b) The solution to the state equation is determined by the statetransition matrix (3.73). How could this matrix function becomputed for the system in (a)?

3.19 Properties of state transition matrices: the concept of a state transi-tion matrix extends to time-varying dynamic systems [3.14]. sup-


pose a dynamic system satisfies x(t) =A( t)x(t), where x(t,) is givenand A(t) is an n X n matrix. We can express the solution in the formx(t) = Q(t, te)x( to). We refer to the n x n matrix @(to, t) as the statetransition matrix. The state transition matrix has the followingproperties:

(b) @(to,tl)@(tl,t~=@(t,,t~ for all t,, t,, and t2;(4 W,,t,)-‘=W,J,);(d) If A(t)&A(s)dr= &A(s)dsA(t), then @(t, to) = exp &A(s)ds

(see P&C 4.29);(e) det @( t, to) = exp I:, trace[A(s)]ds, where trace[A(s)] is the sum

of the diagonal elements of A(s).3.20 A certain system can be represented by a differential equation of

the form i + a$ + a,f = u. The values of the coefficients a, and a2are unknown. However, we have observed the response of theundriven system (u(t) = 0 for t > 0) with various initial conditions.In particular, for f(0) = 1 and i(O) = 0, we find that f(t) = 2e-‘- ee2*and i(t)= 2(eS2’- e-’ ) for t > 0. Also, for f(0) = 0 and i<O) = 1, we findthat a n d i(t)=2ew2’-e-’ f o r t>O.(a) Determine the state equation in terms of a, and a2.(b) Use the transient measurements to determine the state transi-

tion matrix and the precise inverse of the state equation.

3.21 Discrete-time state equations: by using finite-difference approxima-tions for derivatives, an arbitrary nth-order linear constant-coefficient differential equation with initial conditions can beapproximated by an nth-order linear constant-coefficient differenceequation of the form

f((k+n)T)+a,f((k+n-1)7)+*** +a,f(kr)=u(kr)

for k = 0, 1,2,. . . , with f(O),f(T), . . . ,f((n - 1)~) given. The quantity 7is the time increment used in the finite-difference approximation.(a) P u t t h i s nth-order difference equation in state-space form;

that is, develop an equivalent first-order vector differenceequation.

(b) Determine the form of the inverse of the discrete-time stateequation.

3.6 References

[3.1] Bergman, Stefan and M. Schiffer, Kernel Functions and Ellipticin Mathematical Physics, Academic Press, New York, 1953.

Differential Equations

Sec. 3.6 References 143

[3.2]

[3.3]

*[3.4]

[3.5]

[3.6]

[3.7]

[3.8]

[3.9]

[3.10]

*[3.11]

[3.12]

[3.13]

[3.14]

DeRusso, Paul M., Rob J. Roy, and Charles M. Close, State Variables for Engineers,Wiley, New York, 1966.

Forsythe, George E. and Wolfgang R. Wasow, Finite-Difference Methods for PartialDifferential Equations, Wiley, New York, 1960.

Friedman, Bernard, Principles and Techniques of Applied Mathematics, Wiley, NewYork, 1966.

Greenberg, Michael D., Applications of Green’s Functions in Science and Engineering,Prentice-Hall, Englewood Cliffs, N.J., 1971.

Ince, E. L., Ordinary Differential Equations, Dover, New York, 1956.

Kaplan, Wilfred, Advanced Calculus, Addison-Wesley, Reading, Mass., 1959.

Morse, Philip M. and Herman Feshbach, Methods of Theoretical Physics, Parts I andII, McGraw-Hill, New York, 1953.

Ralston, Anthony, A First Course in Numerical Analysis, McGraw-Hill, New York,1965.

Schwartz, L., Théorie des Distributions, Vols. 1 and 2, Hermann & Cie, Paris, 1951,1957.

Stakgold, Ivar, Boundary Value Problems of Mathematical Physics, Volume I, Mac-millan, New York, 1968.

Varga, Richard S., Matrix Iterative Analysis, Prentice-Hall, Englewood Cliffs, N.J.,1962.

Wylie, C. R., Jr., Advanced Engineering Mathematics, 3rd ed., McGraw-Hill, NewYork, 1966.

Zadeh, Lofti A. and Charles A. Desoer, Linear System Theory, McGraw-Hill, NewYork, 1963.

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